SHM of Simple Pendulum Questions in English

(A) In an evacuated chamber,there is no air resistance or drag force acting on the bob of the simple pendulum.
Since the only force acting on the bob is gravity (which is a conservative force),there is no mechanism to dissipate the mechanical energy of the system.
According to the law of conservation of energy,the total mechanical energy remains constant.
Therefore,the amplitude of the oscillation remains constant over time.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.
When the lift is at rest,the effective acceleration due to gravity is $g_{eff} = g$.
So,$T = 2 \pi \sqrt{\frac{l}{g}}$ ...$(i)$
When the lift moves upwards with an acceleration $a = 3g$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + 3g = 4g$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2 \pi \sqrt{\frac{l}{4g}}$.
$T^{\prime} = \frac{1}{2} \times 2 \pi \sqrt{\frac{l}{g}}$.
Substituting the value from Eq. $(i)$,we get $T^{\prime} = \frac{T}{2}$.

(B) For a simple pendulum of length $L$,the time period is given by $T_1 = 2 \pi \sqrt{\frac{L}{g}}$.
For a uniform rod of length $L$ suspended from one end,the moment of inertia about the pivot is $I = \frac{1}{3} mL^2$. The distance of the center of mass from the pivot is $r_{cm} = \frac{L}{2}$.
The time period of a physical pendulum is $T = 2 \pi \sqrt{\frac{I}{mg r_{cm}}}$.
Substituting the values for the rod: $T_2 = 2 \pi \sqrt{\frac{\frac{1}{3} mL^2}{mg (L/2)}} = 2 \pi \sqrt{\frac{2L}{3g}}$.
Now,calculating the ratio $\frac{T_1}{T_2} = \frac{2 \pi \sqrt{L/g}}{2 \pi \sqrt{2L/3g}} = \sqrt{\frac{L/g}{2L/3g}} = \sqrt{\frac{3}{2}}$.

(A) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(PE)$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.
Here,$A = 10 \ cm$ and $x = 4 \ cm$.
The ratio of kinetic energy to potential energy is $\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.
Substituting the values: $\frac{KE}{PE} = \frac{10^2 - 4^2}{4^2} = \frac{100 - 16}{16} = \frac{84}{16}$.
Calculating the result: $\frac{84}{16} = 5.25$.

(A) For a simple pendulum,the tension $(T)$ in the string is given by $T = mg \cos \theta + \frac{mv^2}{l}$.
For small oscillations,$\theta$ is small,so $\cos \theta \approx 1 - \frac{\theta^2}{2}$.
The velocity $v$ is related to the angular displacement $\theta$ by $v = l \frac{d\theta}{dt}$.
Since the pendulum executes $SHM$,$\theta = \theta_0 \sin(\omega t + \phi)$.
Substituting these,we find that $T$ varies with time as $T \propto \cos(2\omega t + 2\phi)$.
The frequency of the tension variation is twice the frequency of the pendulum's oscillation.
Since the bob is not at the mean position at $t=0$,the phase $\phi \neq 0$,and the tension will not be at its maximum or minimum value at $t=0$. Graph $(a)$ represents a periodic variation of tension with time,which is consistent with the physical behavior of a pendulum.

(B) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the pendulum is immersed in a liquid of density $\rho$ and the bob has a density $\sigma$,the effective acceleration due to gravity $g'$ is given by $g' = g(1 - \frac{\rho}{\sigma})$.
Here,$\rho = 1000 \ kg \ m^{-3}$ and $\sigma = \frac{5000}{3} \ kg \ m^{-3}$.
Thus,$g' = g(1 - \frac{1000}{5000/3}) = g(1 - \frac{3000}{5000}) = g(1 - \frac{3}{5}) = g(\frac{2}{5})$.
The time period in water is $t = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(2/5)}} = \sqrt{\frac{5}{2}} \cdot 2\pi \sqrt{\frac{l}{g}} = \sqrt{\frac{5}{2}} T$.
Therefore,$\frac{T}{t} = \sqrt{\frac{2}{5}}$.

(A) When the hydrometer is floating in equilibrium,its weight is balanced by the buoyant force.
When it is pushed down by a small distance $x$,the additional buoyant force acts as the restoring force.
The additional volume submerged is $V = \pi r^2 x$.
The additional buoyant force is $F = \rho V g = \rho (\pi r^2 x) g$.
Since this force acts in the direction opposite to the displacement,the restoring force is $F_{restoring} = -\rho \pi r^2 g x$.
Using Newton's second law,$m a = -\rho \pi r^2 g x$,which gives $a = -(\frac{\rho \pi r^2 g}{m}) x$.
Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{\rho \pi r^2 g}{m}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{\rho \pi r^2 g}}$.

(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length of the pendulum measured from the point of suspension to the center of mass of the bob.
Initially,the center of mass of the water-filled sphere is at its geometric center.
As water drains out from the hole at the bottom,the center of mass of the system shifts downwards. This increases the effective length $l$,causing the time period $T$ to increase.
As the water level continues to drop,the center of mass reaches its lowest point and then begins to rise back towards the geometric center of the sphere as the bob becomes empty.
Once the water is completely drained,the center of mass returns to the geometric center of the hollow sphere,restoring the original effective length $l$ and thus the original time period $T$.

(A) Let $T_1$ be the time period when the lift moves upwards with acceleration $a$, and $T_2$ be the time period when the lift moves downwards with the same acceleration $a$.
The effective acceleration due to gravity when moving upwards is $g' = g + a$, and when moving downwards is $g'' = g - a$.
The time periods are given by:
$T_1 = 2\pi \sqrt{\frac{l}{g+a}}$ $(i)$
$T_2 = 2\pi \sqrt{\frac{l}{g-a}}$ $(ii)$
Given the ratio $T_1 : T_2 = 1 : 2$, we have:
$\frac{T_1}{T_2} = \sqrt{\frac{g-a}{g+a}} = \frac{1}{2}$
Squaring both sides:
$\frac{g-a}{g+a} = \frac{1}{4}$
$4(g - a) = g + a$
$4g - 4a = g + a$
$3g = 5a$
$a = \frac{3g}{5} = \frac{3 \times 10}{5} = 6 \,ms^{-2}$

(A) The initial potential energy of the bob at the horizontal position is $PE_i = mgh$,where $h = L = 200 \ cm = 2 \ m$.
So,$PE_i = m \times 10 \times 2 = 20mg$.
When the bob reaches the mean position,its potential energy is converted into kinetic energy.
Given that $10 \%$ of the initial energy is lost,the remaining energy is $90 \%$ of the initial potential energy.
$KE_f = 0.9 \times PE_i = 0.9 \times 20mg = 18mg$.
At the mean position,$KE_f = \frac{1}{2}mv^2$.
Equating the two: $\frac{1}{2}mv^2 = 18mg$.
$v^2 = 36g = 36 \times 10 = 360$.
$v = \sqrt{360} \approx 18.97 \ m \ s^{-1}$.
Wait,re-evaluating: $PE_i = mgL$. $KE_f = 0.9 \times mgL = \frac{1}{2}mv^2$.
$v^2 = 1.8 \times g \times L = 1.8 \times 10 \times 2 = 36$.
$v = 6 \ m \ s^{-1}$.

(C) Given: $h = 20 \,m$, $v = 31.4 \,ms^{-1}$.
Using the equation of motion $v^2 = u^2 + 2gh$ where $u = 0$:
$v^2 = 2gh$
$g = \frac{v^2}{2h} = \frac{31.4 \times 31.4}{2 \times 20} = \frac{985.96}{40} \approx 24.649 \,ms^{-2}$.
Note that $31.4 \approx 10\pi$, so $g = \frac{(10\pi)^2}{40} = \frac{100\pi^2}{40} = 2.5\pi^2 \,ms^{-2}$.
$A$ seconds pendulum has a time period $T = 2 \,s$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides: $T^2 = 4\pi^2 \frac{l}{g} \Rightarrow l = \frac{T^2 g}{4\pi^2}$.
Substituting $T = 2 \,s$ and $g = 2.5\pi^2 \,ms^{-2}$:
$l = \frac{2^2 \times 2.5\pi^2}{4\pi^2} = \frac{4 \times 2.5\pi^2}{4\pi^2} = 2.5 \,m$.

(B) The time period of a simple pendulum in air is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
When the pendulum oscillates in a liquid of density $\sigma$,the effective acceleration due to gravity $g'$ is given by $g' = g \left(1 - \frac{\sigma}{\rho}\right)$,where $\rho$ is the density of the bob.
The time period in water is $T' = 2 \pi \sqrt{\frac{L}{g'}}$.
Given $T' = \sqrt{2} T$,we have $\sqrt{2} = \frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g(1 - \sigma/\rho)}} = \frac{1}{\sqrt{1 - \sigma/\rho}}$.
Squaring both sides,$2 = \frac{1}{1 - \sigma/\rho}$,which implies $1 - \frac{\sigma}{\rho} = \frac{1}{2}$.
Thus,$\frac{\sigma}{\rho} = \frac{1}{2}$.
Since the density of water $\sigma = 1 \text{ g/cm}^3$,the relative density of the bob is $\rho = 2$.

(C) The bob of the simple pendulum experiences both the acceleration due to gravity $(g)$ and the centripetal acceleration $(a_c = v^2/R)$ due to the circular motion of the car.
These two accelerations are perpendicular to each other.
The effective acceleration $(g')$ acting on the bob is given by $g' = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + (v^2/R)^2} = \sqrt{g^2 + v^4/R^2}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{L/g'}$.
Substituting the values: $T = 2\pi \sqrt{L / \sqrt{g^2 + v^4/R^2}} = 2\pi \sqrt{1 / (g^2 + v^4/R^2)^{1/2}} = 2\pi / (g^2 + v^4/R^2)^{1/4}$.
Comparing this with $T = 2\pi / \alpha^{1/4}$,we get $\alpha = g^2 + v^4/R^2$.
Given $g = 10 \ m/s^2$,$v = 10 \ m/s$,and $R = 100 \ m$:
$\alpha = (10)^2 + (10)^4 / (100)^2 = 100 + 10000 / 10000 = 100 + 1 = 101$.

(A) When a small sphere of radius $r$ rolls without slipping on a concave surface of radius $R$,the effective radius of the path of the center of mass is $(R - r)$.
The torque about the point of contact is $\tau = -mg(R-r)\sin\theta \approx -mg(R-r)\theta$.
The moment of inertia of the sphere about the point of contact is $I = I_{cm} + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$.
The equation of motion is $\tau = I\alpha$,so $-mg(R-r)\theta = \frac{7}{5}mr^2 \frac{d^2\theta}{dt^2}$.
This gives $\frac{d^2\theta}{dt^2} = -\frac{5g(R-r)}{7r^2}\theta$.
However,if the sphere is assumed to slide without friction,the motion is equivalent to a simple pendulum of length $l = (R-r)$.
Given $r << R$,the effective length $l \approx R$.
The time period is $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{R}{g}}$.

(D) Given: Length of simple pendulum $L = 1 \,m$,acceleration of elevator $a = 2 \,m/s^2$,and acceleration due to gravity $g = 10 \,m/s^2$.
When an elevator moves upward with an acceleration $a$,the effective acceleration due to gravity $g_{eff}$ is given by $g_{eff} = g + a$.
Substituting the values,$g_{eff} = 10 + 2 = 12 \,m/s^2$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
Substituting the values,$T = 2\pi \sqrt{\frac{1}{12}} = 2\pi \frac{1}{\sqrt{4 \times 3}} = 2\pi \frac{1}{2\sqrt{3}} = \frac{\pi}{\sqrt{3}} \,s$.

(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$,so $T^2 = 4 \pi^2 \left(\frac{l}{g}\right) \quad ...(i)$
When the length is increased by $10 \ cm$,the new time period $T_1$ is given by $T_1^2 = 4 \pi^2 \left(\frac{l+10}{g}\right) \quad ...(ii)$
When the length is decreased by $10 \ cm$,the new time period $T_2$ is given by $T_2^2 = 4 \pi^2 \left(\frac{l-10}{g}\right) \quad ...(iii)$
Adding equations $(ii)$ and $(iii)$,we get:
$T_1^2 + T_2^2 = 4 \pi^2 \left(\frac{l+10}{g}\right) + 4 \pi^2 \left(\frac{l-10}{g}\right)$
$T_1^2 + T_2^2 = \frac{4 \pi^2}{g} (l + 10 + l - 10)$
$T_1^2 + T_2^2 = \frac{4 \pi^2}{g} (2l)$
$T_1^2 + T_2^2 = 2 \left(4 \pi^2 \frac{l}{g}\right)$
Substituting equation $(i)$ into this,we get:
$T_1^2 + T_2^2 = 2 T^2$

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
The frequency $f$ (number of oscillations per unit time) is $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.
Since $g$ is constant at a given place,$f \propto \frac{1}{\sqrt{L}}$,which implies $L \propto \frac{1}{f^2}$.
Let the initial frequency be $f_1 = 72 \text{ oscillations/min}$ and the final frequency be $f_2 = 90 \text{ oscillations/min}$.
Then,$\frac{L_2}{L_1} = \left( \frac{f_1}{f_2} \right)^2 = \left( \frac{72}{90} \right)^2 = \left( \frac{4}{5} \right)^2 = \frac{16}{25} = 0.64$.
This means the new length $L_2$ is $64 \%$ of the original length $L_1$.
The decrease in length is $\Delta L = L_1 - L_2 = L_1 - 0.64 L_1 = 0.36 L_1$.
Therefore,the percentage decrease is $36 \%$.

(C) The forces acting on the sphere are the gravitational force $mg$ (downwards) and the electric force $qE$ (upwards,since the field is vertically upwards and charge is positive).
The net downward force $F$ is given by $F = mg - qE$.
We can write this as $F = m(g - \frac{qE}{m})$.
Thus,the effective acceleration due to gravity $g'$ is $g' = g - \frac{qE}{m}$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g'}}$.
Substituting the value of $g'$,we get $T = 2\pi \sqrt{\frac{l}{g - \frac{qE}{m}}}$.

(B) Given, length of the rod, $l = 1.8 \,m$.
The time period of oscillation of a uniform rod of length $l$ pivoted at one end is given by $T = 2 \pi \sqrt{\frac{I}{mgR}}$, where $I$ is the moment of inertia about the pivot and $R$ is the distance from the pivot to the center of mass.
For a rod pivoted at one end, $I = \frac{ml^2}{3}$ and $R = \frac{l}{2}$.
Substituting these values into the formula:
$T = 2 \pi \sqrt{\frac{ml^2/3}{mg(l/2)}} = 2 \pi \sqrt{\frac{2l}{3g}} = 2 \pi \sqrt{\frac{2 \times 1.8}{3g}} = 2 \pi \sqrt{\frac{1.2}{g}}$.
The time period of a simple pendulum of length $l'$ is $T' = 2 \pi \sqrt{\frac{l'}{g}}$.
Since $T = T'$, we have $2 \pi \sqrt{\frac{l'}{g}} = 2 \pi \sqrt{\frac{1.2}{g}}$.
Squaring both sides, we get $l' = 1.2 \,m$.

(C) The bob of the simple pendulum is hanging vertically below a fixed bob. Both bobs carry an identical charge $q$.
Since the bobs are aligned vertically,the electrostatic force of repulsion acts along the line joining the two bobs,which is the same line as the string.
This electrostatic force acts in the upward direction on the moving bob,opposing the gravitational force (weight $mg$).
However,for a simple pendulum,the restoring force is provided by the component of gravity perpendicular to the string,which is $mg \sin \theta$.
The electrostatic force acts along the string and does not have a component perpendicular to the string.
Therefore,the electrostatic force does not affect the restoring force or the motion of the pendulum.
Thus,the time period of the pendulum remains unchanged: $T = 2 \pi \sqrt{\frac{l}{g}}$.

(A) Let $l$ be the length of the string. At the extreme position,the velocity is zero,so the tension is $T_{\min } = mg \cos \theta$,where $\theta$ is the maximum angular displacement.
At the lowest point (mean position),the tension is maximum,given by $T_{\max } = mg + \frac{mv^2}{l}$.
By the law of conservation of energy between the extreme position and the mean position: $\frac{1}{2} mv^2 = mgl(1 - \cos \theta)$,which implies $v^2 = 2gl(1 - \cos \theta)$.
Substituting $v^2$ into the expression for $T_{\max }$: $T_{\max } = mg + \frac{m(2gl(1 - \cos \theta))}{l} = mg + 2mg(1 - \cos \theta) = mg(1 + 2 - 2 \cos \theta) = mg(3 - 2 \cos \theta)$.
Given $T_{\max } = 2 T_{\min }$,we have $mg(3 - 2 \cos \theta) = 2(mg \cos \theta)$.
$3 - 2 \cos \theta = 2 \cos \theta \implies 4 \cos \theta = 3 \implies \cos \theta = \frac{3}{4}$.
Substituting $\cos \theta = \frac{3}{4}$ into the expression for $T_{\max }$: $T_{\max } = 2mg \cos \theta = 2mg \left(\frac{3}{4}\right) = \frac{3mg}{2}$.

(B) For a bob of mass $m$ moving in a vertical circular path of radius $L$ with speed $v$,the forces acting on the bob along the radial direction are the tension $T$ (towards the center) and the component of weight $mg \cos \theta$ (away from the center).
The net centripetal force required for circular motion is provided by the resultant of these radial forces:
$T - mg \cos \theta = \frac{mv^2}{L}$
Rearranging the equation to solve for the tension $T$:
$T = \frac{mv^2}{L} + mg \cos \theta$

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
At a height $h = R$ (where $R$ is the radius of the Earth),the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2 = g \left( \frac{R}{R+R} \right)^2 = \frac{g}{4}$.
Given $g = \pi^2 \ m/s^2$ at the surface,the effective gravity at height $h$ is $g' = \frac{\pi^2}{4} \ m/s^2$.
Substituting the values $\ell = 4.0 \ m$ and $g' = \frac{\pi^2}{4} \ m/s^2$ into the time period formula:
$T = 2\pi \sqrt{\frac{4}{\pi^2 / 4}} = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \times \frac{4}{\pi} = 8 \ s$.

(C) The electric field $E$ produced by an infinite plane sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2 \varepsilon_{0}}$.
Since the bob carries a charge $q$,it experiences an upward electrostatic force $F_e = qE = \frac{q \sigma}{2 \varepsilon_{0}}$.
The effective acceleration due to gravity $g_{\text{eff}}$ acting on the bob is given by $m g_{\text{eff}} = mg - F_e = mg - \frac{q \sigma}{2 \varepsilon_{0}}$.
Thus,$g_{\text{eff}} = g - \frac{q \sigma}{2 \varepsilon_{0} m}$.
The time period $T$ of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$.
Substituting the value of $g_{\text{eff}}$,we get $T = 2 \pi \sqrt{\frac{L}{g - \frac{q \sigma}{2 \varepsilon_{0} m}}}$.

(C) The time period of the pendulum is $T = 2 \,s$. The angular frequency is $\omega = \frac{2 \pi}{T} = \pi \,rad/s$.
At the lowest point,the displacement is $x = 0$,so the equation of motion is $x(t) = A \sin(\omega t)$,where $A$ is the amplitude.
The velocity at any time $t$ is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.
At $t = 0$,$v(0) = v_{0} = A \omega$,so $A = \frac{v_{0}}{\omega}$.
We need the height $h$ at $t = 2.25 \,s$. Since $T = 2 \,s$,$t = 2.25 \,s = T + 0.25 \,s = T + \frac{T}{8}$.
At $t = \frac{T}{8} = 0.25 \,s$,the velocity is $v = v_{0} \cos(\omega \cdot \frac{T}{8}) = v_{0} \cos(\frac{2 \pi}{T} \cdot \frac{T}{8}) = v_{0} \cos(\frac{\pi}{4}) = \frac{v_{0}}{\sqrt{2}}$.
Using the principle of conservation of mechanical energy: $\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + mgh$.
Substituting $v = \frac{v_{0}}{\sqrt{2}}$,we get $\frac{1}{2} v_{0}^{2} = \frac{1}{2} (\frac{v_{0}}{\sqrt{2}})^{2} + gh$.
$\frac{1}{2} v_{0}^{2} = \frac{1}{4} v_{0}^{2} + gh$.
$gh = \frac{1}{4} v_{0}^{2} \implies h = \frac{v_{0}^{2}}{4g}$.

(B) The time period $T$ of a simple pendulum is given by the formula: $T = 2\pi \sqrt{\frac{L}{g}}$.
Squaring both sides,we get: $T^2 = \frac{4\pi^2 L}{g}$.
Taking the reciprocal of both sides,we get: $\frac{1}{T^2} = \frac{g}{4\pi^2 L}$.
This equation is of the form $y = \frac{k}{x}$,where $y = \frac{1}{T^2}$,$x = L$,and $k = \frac{g}{4\pi^2}$ is a constant.
This represents a rectangular hyperbola,which corresponds to the plot shown in option $B$.

(C) The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $T \propto \sqrt{\ell}$.
Given that the time duration $t$ is constant,the number of oscillations $n$ is inversely proportional to the time period $T$ $(n = \frac{t}{T})$.
Therefore,$n \propto \frac{1}{\sqrt{\ell}}$,which means $n^2 \propto \frac{1}{\ell}$ or $\ell \propto \frac{1}{n^2}$.
Initially,$n_1 = 20$ and $\ell_1 = 30 \ cm$.
Finally,$n_2 = 40$.
Using the ratio: $\frac{\ell_2}{\ell_1} = \left( \frac{n_1}{n_2} \right)^2$.
$\ell_2 = 30 \times \left( \frac{20}{40} \right)^2 = 30 \times \left( \frac{1}{2} \right)^2 = 30 \times \frac{1}{4} = 7.5 \ cm$.

(D) The total mechanical energy $E$ of a simple pendulum is the sum of its kinetic energy and potential energy,which remains constant throughout the motion.
Given total energy $E = 0.02 \text{ J}$.
At the equilibrium position (mean position),the potential energy of the bob is zero,so the total energy is entirely in the form of kinetic energy.
Therefore,$E = K.E_{max} = \frac{1}{2} m v_{max}^2$.
Given mass $m = 20 \text{ g} = 0.02 \text{ kg}$.
Substituting the values into the equation:
$0.02 = \frac{1}{2} \times 0.02 \times v_{max}^2$
$1 = \frac{1}{2} v_{max}^2$
$v_{max}^2 = 2$
$v_{max} = \sqrt{2} \approx 1.414 \text{ m/s}$.
Thus,the speed at the equilibrium position is approximately $1.41 \text{ m/s}$.

(B) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$.
Given that the time taken for $30$ oscillations is $60 \text{ s}$, the time period $T$ is calculated as $T = \frac{60}{30} = 2 \text{ s}$.
Substituting the values into the formula: $2 = 2\pi \sqrt{\frac{L}{9.8}}$.
Squaring both sides of the equation, we get $4 = 4\pi^2 \frac{L}{9.8}$.
Given $\pi^2 = 9.8$, we substitute this into the equation: $4 = 4 \times 9.8 \times \frac{L}{9.8}$.
Simplifying the expression, we get $4 = 4L$, which gives $L = 1 \text{ m}$.
Therefore, the correct option is $B$.